Center for a subgroup of $\operatorname{GL}_{2}(\mathbb{F_{7}})$

Question

I have a subgroup of $\operatorname{GL}_{2}(\mathbb{F_{7}})$ being

$G= \left\{ \begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} \text{with $\bar{a}$ and $\bar{c}$ in $\mathbb{F}^{*}_{7}$ and $\bar{b}$ in $\mathbb{F}_{7}$} \right\} $ and I have to compute the center of it.

The definition according the book for the center is

$Z(G) = \left\{ g \in G \mid xg=gx \text{ for all $x \in G$} \right\}$, thus in words: all elements of $G$ that commute with all other elements of $G$.

My plan was to take a $g \in G$ with $g = \begin{bmatrix} \bar{d} & \bar{e} \\ \bar{0} & \bar{f} \end{bmatrix} \text{with $\bar{d}$ and $\bar{f}$ in $\mathbb{F}^{*}_{7}$ and $\bar{e}$ in $\mathbb{F}_{7}$}. $

Then $\begin{bmatrix} \bar{d} & \bar{e} \\ \bar{0} & \bar{f} \end{bmatrix} \cdot \begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} = \begin{bmatrix} \bar{d}\bar{a} & \bar{d}\bar{b}+\bar{e}\bar{c} \\ \bar{0} & \bar{f}\bar{c} \end{bmatrix}$

and likewise

$\begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} \cdot \begin{bmatrix} \bar{d} & \bar{e} \\ \bar{0} & \bar{f} \end{bmatrix} = \begin{bmatrix} \bar{a}\bar{d} & \bar{a}\bar{e}+\bar{b}\bar{f} \\ \bar{0} & \bar{c}\bar{f} \end{bmatrix}$.

Thus I would say $\begin{bmatrix} \bar{a}\bar{d} & \bar{a}\bar{e}+\bar{b}\bar{f} \\ \bar{0} & \bar{c}\bar{f} \end{bmatrix} = \begin{bmatrix} \bar{d}\bar{a} & \bar{d}\bar{b}+\bar{e}\bar{c} \\ \bar{0} & \bar{f}\bar{c} \end{bmatrix}$, but I don't know how to make sense of this. Am I right and if so: how to proceed?

Answer 1

Now you know that $a,b,c,d,e,f$ are all in $\mathbb{F}_{7}$.

The diagonal entries are clearly equal, so you need to also have $ae+bf = db+ec$ for every $d$,$e$, $f$. Your matrices are invertible, so $a$ and $b$ are both nonzero (as are $d$ and $f$). Also, you can choose some specific values of $d$, $e$, and $f$ that will impose some necessary conditions on $a$, $b$, $c$ (try $d=e=f=1$ for example).

Answer 2

In the matrix equation, you may choose $d,e,f$ (subject to $df \ne 0)$ to derive conditions for $a,b,c$.

Start by choosing $ e = 0$ and $d \ne f$. What does that imply for $a,b,c$?

Next choose $e = 1$. What does that imply?

These two steps result in a very special form of matrices $\pmatrix{a & b \\ 0 & c}$. Now show that all these matrices indeed commute with all matrices in $G$, and you have solved the problem.